"二叉树运算求值：根据节点颜色进行乘法或加法计算"

解析：首先需要定义一个二叉树节点类，包含值、左子节点和右子节点。然后根据输入的节点个数n,创建一个二叉树。接着根据输入的根节点和颜色信息，构建二叉树。最后计算根节点的值并输出。 代码如下： ```c #include <stdio.h> #include <stdlib.h> typedef struct TreeNode { int val; struct TreeNode *left; struct TreeNode *right; } TreeNode; TreeNode* createNode(int val) { TreeNode *node = (TreeNode *)malloc(sizeof(TreeNode)); node->val = val; node->left = NULL; node->right = NULL; return node; } void buildTree(TreeNode **root, int *p, int n, int c[]) { if (p[0] == -1) return; (*root) = createNode(p[0]); int left_count = 0, right_count = 0; for (int i = 1; i < n; i++) { if (p[i] == p[0]) left_count++; else if (p[i] == p[0]) right_count++; } int left_val = c[p[0]]; int right_val = c[p[0]]; int left_index = -1, right_index = -1; for (int i = 1; i < n; i++) { if (c[p[i]] == left_val) left_index = i; else if (c[p[i]] == right_val) right_index = i; } int left_parent = p[0], right_parent = p[0]; int left_color = c[left_parent], right_color = c[right_parent]; int left_child = p[left_index], right_child = p[right_index]; (*root)->left = buildTree(&((*root)->left), p + left_count, n, c); (*root)->right = buildTree(&((*root)->right), p + left_count + right_count, n, c); } int getRootValue(TreeNode *root) { if (root == NULL) return 0; int left_val = getRootValue(root->left); int right_val = getRootValue(root->right); int color = root->left != NULL && root->right != NULL ? "red" : "blue"; int result = root->val; if (color == "red") result *= left_val; else result += right_val; return result; } int main() { int n; scanf("%d", &n); int *p = (int *)malloc(n * sizeof(int)); int *c = (int *)malloc(n * sizeof(int)); p[0] = n > 1 ? scanf("%d", &p[1]) : -1; c[n > 1] = scanf("%d", &c[n > 1]); p++; c++; memset(p, -1, (n + 1) * sizeof(int)); memset(c, 0, n * sizeof(int)); int root_value = getRootValue(createNode(0)); printf("%d ", root_value); free(p); free(c); free(createNode(0)); return 0; } ```